29 percent of the employees of a Fortune 500 company are unhappy. A random sample of 9 employees is selected. What is the probability that the sample contains exactly 3 unhappy employees?
Solution
Given that ,
p = 0.20
1 - p = 1 - 0.20 = 0.80
n = 9
Using binomial probability formula ,
P(X = x) = (n C x) * px * (1 - p)n - x
P(X = 3) = (9 C 3) * (0.20)3 * (0.80)6
= 0.176160
Probability = 0.1762
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